Question

My question was: A saturated solution of MgF2 at 23 degrees C was prepared by dissolving solid MgF2 in 1 L water. The Ksp of MgF2 is 1.5e-5.
a.)Calculate the mass of MgF2 dissolved.
b.) When .1 mols of solid KF was dissolved in the MgF2 solution, precipitation was observed. Find the mass of the additional precipitate.
I think I get a, I just found the molarity of Mg2+ using the Ksp and then converted to grams; but, I have no idea how to do b. Thanks in advance.

Your answer was: The b part is a common ion problem; i.e., F^- is the ion common to both KF and MgF2.
For part b, Ksp = (Mg^2+)(F^-)^2
The common ion shifts the solubility to the left. MgF2 ==> Mg^2+ + 2F^-
because of Le Chatlier's Principle. The new solubility will = new (Mg^2+) and you can convert that to g MgF2 in the 1L. Subtraction from the initial amount will give you the amount pptd.

New question: I'm sorry but I am still so confused on pretty much everything. How do you calculate the new solubility? And how do you convert that to g MgF2 in 1L? And what is the initial amount that you subtract from?

Answer 1

You said you understood part a. That's the initial amount of MgF2 dissolved in the 1L of water. You should have done this for part a.
............MgF2 ==> Mg^2+ + 2F^-
I...........solid.....0.......0
C...........solid.....x.......2x
E...........solid.....x.......2x

Ksp = (Mg^2+)(F^-)^2
1.5E-5 = (x)(2x)^2 = 4x^3
x = solubility MgF2 = about 0.016 but you should do it a little more accurately. That works out to approx 0.97grams MgF2/L

For part b you have two things to consider. The F from MgF2 and the F from KF.
(Mg^2+) from MgF2 = x
(F^-) = 0.032 from MgF2 and 0.1 from KF You may want to assume that 0.1 + 0.032 = 0.1 (the easy way but not too accurate) but at least do this first to see what the answer is.
(x)(0.1)^2 = 1.5E-5
x = (Mg^2+) = (MgF2) = about 1.5E-3 M which converts to 0.0934g MgF2. You had about 0.97 dissolved form part a so the difference is what ppts in part b.
I think the assumption is not valid and I believe you should go through the quadratic formula to find (Mg^2+)=(Mg^2+) as above. That quadratic on second thought it isn't a quadratic but a cubic) you get from (Mg^2+)(F^-)^2 = Ksp
1.5E-5 = (x)(2x+0.1)^2
Solve for x, convert to grams/L, subtract from your number from part a and that will be the amount pptd.
Th whole point of these problems is to show you that by adding a common ion you can decrease the solubility of a sparingly soluble salt dramatically.
You can find calculators on the web that will solve a cubic or you can use that assumption (modified of course) so that 0.1 + 0.032 = 0.132 and use that instead of the 0.1 I used above. Hope this helps.